CHAPTER I

APPLICATION OF CIRCUIT LAWS

2

Introduction

Generally, we require 3 steps to analyzeAC Circuit Transform the circuit to the phasor /

frequency domain Solve the problem using any

technique Transform the resulting phasor /

frequency domain to the time domain expression.

3

Time domain to Phasor

4

Impedance in Frequency Domain

ELEMENT

IMPEDANCE FORMULAPHASOR FORM

RECTANGULAR FORM

R ZR R R 0 R + j0

L ZL jωL XL 90 0 + jXL

C ZC 1/jωC XC -90 0 - jXC

5

Methods of solving

Kirchhoff’s Voltage Law (KVL) Kirchhoff’s Current Law (KCL) Voltage Divider Rule (VDR) Current Divider Rule (CDR) Star / Delta Transformation ( / )

6

Kirchhoff’s Voltage Law (KVL)

Algebraic sum of voltage drops around closed loop is zero

V = 0Voltage drop = voltage rise

V = V1+ V2 + … + VN

= I (Z1 + Z2 + … + ZN)

7

Kirchhoff’s Current Law (KCL)

Algebraic sum of current at any node is zero

I = 0Current In = Current Out

I = I1+ I2 + … + IN

= V (1 / Z1 + 1 / Z2 + … + 1 / ZN)

8

Voltage Divider Rule (VDR)

VZZ

ZV

21

11

VZZ

ZV

21

22

9

Current Divider Rule (CDR)

IZZ

ZI

21

21

IZZ

ZI

21

12

10

Star / Delta ( / ) Transformation

1

313221a Z

ZZZZZZZ

2

313221b Z

ZZZZZZZ

- Conversion

3

313221C Z

ZZZZZZZ

11

Delta / Star ( / Y) Transformation

cba

cb1 ZZZ

ZZZ

cba

ca2 ZZZ

ZZZ

- Y Conversion

cba

ba3 ZZZ

ZZZ

12

Example 1Find the input impedance, Zin of the circuit

(ω=50rad/s)

13

Solution Example 1

j10Ω)10j(50)(2

1

jω

1Z

31

C

Let

Z1 = Impedance of the 2mF capacitor

Z2 = Impedance of the 3 resistor in series with the 10mF capacitor

Z3 = Impedance of the 0.2H inductor in series with 8 resistor

Then

j2)Ω(3)10j(50)(10

13

jω

13Z

32

C

j10)Ω(8j(50)(0.2)8jω8Z3 L

14

Solution Example 1

The input impedance is

j811

j10)j2)(8(3j10Z||ZZZ 321in

j1.073.22j10

)07.1122.3( j

15

Example 2Find the input impedance, Zin of the circuit (ω=10rad/s)

16

Solution Example 2

j50)Ω20()10j(10)(2

120

jω

120Z

31

C

Let

Z1 = Impedance of the 2mF capacitor in series with 20 resistor

Z2 = Impedance of the 4mF capacitor

Z3 = Impedance of the 2H inductor in series with 50 resistor

Then

j25Ω)10j(10)(4

1

jω

1Z

32

C

j20)Ω(50(2))j(1005jω05Z3 L

17

Solution Example 2

The input impedance is

32

321321in ZZ

ZZZZ||ZZZ

j20)50j25

j20)j25(50j5020

j73.76)Ω(32.38

j23.7638.21j5020

18

Example 3Determine Vo (t)

19

Solution example 3

Step 1: Transfer the circuit into frequency domain

20

Solution Example 3

0Ω6Z1

Step 2: Solve the circuit using any method

Let

Z1 = Impedance of the 60 resistor

Z2 = Impedance of the parallel combination of the 10mF capacitor and 5H inductor

Then

j100Ωj20j25-

)(-j25)(j20j20||j25Z2

21

Solution Example 3

By using voltage divider rule

)15(20j10060

j100V

ZZ

ZV o

s21

20

)15)(2030.96(0.8575 oo

V)96.154cos(15.17)( o ttvo

V15.9617.15 o

Step 3: Convert the result to the time domain

22

Example 4Determine Vo (t)

23

Solution Example 4

os 7510V

Step 1: Transfer to the frequency domain

j2)j(10)(1/20

1

jω

1

C

j5j(10)(0.5)jω L

Voltage source

0.5H inductor

(1/20)F capacitor

24

Solution Example 4

j2Z2

Step 2: Solve the circuit using any method

Let

Z1 = Impedance of the 0.5H inductor in parallel with the 10 resistor

Z2 = Impedance of the (1/20)F capacitor

Then

j42j510

(10)(j5)j5||10Z1

25

Solution Example 4

By using voltage divider rule

)57(10j2-j42

j2-V

ZZ

ZV o

s21

20

V)6010cos(071.7)( o ttvo

V60-071.7 o

Step 3: Convert the result to the time domain

26

Example 5Find current I

27

Solution Example 5Step 1: Transform the circuit from delta to star connection ( to Y)

28

Solution Example 5

j3.2Ω10

j2)4(4

10

j4(8)Zbn

j3.2)Ω(1.610

j2)4(4

10

j4)-8(2Zcn

Calculate new impedances after the transformation

j0.8)Ω(1.610

j2)4(4

8j42j4

j4)-j4(2Zan

29

Solution Example 5

The total input impedance is

8)j6(Z||j3)(ZZ12Z cnbnanin

j2.8)(9.6||(j0.2)j0.81.612

Ω4.20413.64j113.6 o

j39.6

j2.8)j0.2(9.6j0.813.6

30

Solution Example 5

The desire current is

o

o

4.20413.64

050

Z

VI

A4.204666.3 o

31

Example 6Find current I

32

Solution Example 6

Step 1: Transform the circuit from

delta to star connection

( to Y)

33

Solution Example 6

j2.82)Ω0.24(j68

j5)j3(8-Zbn

j0.72)Ω(0.9668

j4(-j3)Zcn

j

Calculate new impedances after the transformation

j3.76)Ω(0.32j68

j8)4(-5

j3-j58j4

j5)j4(8Zan

34

Solution Example 6

j2.82)-9.76j1.76(5.32

j2.82)6j1.76)(9.7(5.32Zcn

The total input impedance is

)01(Z||j2)5(ZZZ bnancnin

j2.82)(9.76||j1.76)(5.32Zcn

j0.40743.744j0.7296.0

)3.802-4.714(j0.3126704.4 o

35

Solution Example 6

The desire current is

o

o

3.802714.4

030

Z

VI

A802.3364.6 o